Another Gravitational potential energy problem

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SUMMARY

This discussion focuses on calculating the final velocities of two Jupiter-sized planets as they collide, starting from a distance of 1.30×1011 m apart. The participants analyze the gravitational potential energy and kinetic energy equations, specifically using the conservation of energy principle: Kf + Uf = Ki + Ui. Key insights include the correct interpretation of initial and final distances (ri and rf) and the need to account for both planets' contributions to kinetic energy, leading to the conclusion that the final velocity formula should include a factor of √2.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy concepts
  • Familiarity with the conservation of energy principle in physics
  • Knowledge of the mass and radius of Jupiter (mass: 1.90×1027 kg, radius: 6.99×107 m)
  • Ability to manipulate equations involving square roots and fractions
NEXT STEPS
  • Review the conservation of energy in gravitational systems
  • Learn how to calculate center of mass for two-body systems
  • Study the implications of kinetic energy distribution in multi-body collisions
  • Explore gravitational potential energy calculations for varying distances
USEFUL FOR

Students and educators in physics, particularly those focusing on gravitational interactions, energy conservation, and celestial mechanics. This discussion is beneficial for anyone solving problems related to planetary motion and collisions.

neo982
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Homework Statement


Two Jupiter-size planets are released from rest 1.30×10^11 m apart.

What are their speeds as they crash together?

Homework Equations


The Attempt at a Solution


I have done K_f + U_f = K_i + U_i with K_i being zero because they have crashed together so... K_f + U_f = U_i is the equation I get...when solving for velocity final I get v_f = sqrt(2*G*m*(-1/r_i +1/r_f)) and I am not so sure I am making r_i and r_f the right values, I have made r_i the 1.30×10^11 and the r_f to be 2*radius of Jupiter which is 6.99*10^7 (from book) I keep getting v_f to be 42,566 m/s but that's not right

also from book the mass of Jupiter is 1.90*10^27 kg...
 
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The r's in your problem have nothing to do with the radius of the planet. They are the distance from the center of mass of the system to the two bodies. And they shouldn't have opposite signs.
 
well if you solve 1/2mv^2 + (-Gm1m2/r_f) = (-Gm1m2/r_i) for velocity you get that (-1/r +1/r) part so they are not opposite signs, but anyway...so the initial r would be half of the distance given which is 1/2 (1.3x10^11) = 6.5x10^10 right...if this is true then I am not sure what would be the r final?
 
Hmm. You've got me. You are much more correct than my sloppy thinking. Apologies. Now I'm having some problems figuring out how your answer could be incorrect. So far I'm baffled. But I haven't put numbers in...
 
You have a factor of two missing. If you think in the center of mass system then the central mass is 2*m, then cut your diameters in half to make them radii. The m in the KE remains the same. I get sqrt(2) times what you do. I'm counting on you to figure this out and not make me give an incoherent explanation again.
 
Im not sure if I follow you...what would the v_f eq look like..?
 
neo982 said:

The Attempt at a Solution


I have done K_f + U_f = K_i + U_i with K_i being zero because they have crashed together so... K_f + U_f = U_i is the equation I get...when solving for velocity final I get v_f = sqrt(2*G*m*(-1/r_i +1/r_f)) and I am not so sure I am making r_i and r_f the right values, I have made r_i the 1.30×10^11 and the r_f to be 2*radius of Jupiter which is 6.99*10^7 (from book) I keep getting v_f to be 42,566 m/s but that's not right
Your values for r_i & r_f seem OK to me. But you have an extra sqrt(2) in your formula for v_f. I suspect that comes from you setting KE = 1/2mv^2--but realize that there are two planets sharing that KE.
 
so do I need to make the eq 2*K_f +U_f = U_i with the r's 1.3x10^11 and 2*6.99x10^7 or make them ... half of 1.3x10^11 and the other just 6.99x10^7 ? well...I i know i get the right answer from using 1.3x10^11 and (2*6.99x^10^7) but what is the reason to use the diameter instead of cutting them in half ...?

also, is that the right way to set it up because I get the right answer from my 42,556 / sqrt (2)...or is that the same thing by making 2K
 
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neo982 said:
so do I need to make the eq 2*K_f +U_f = U_i with the r's 1.3x10^11 and 2*6.99x10^7
Right--where K_f is the kinetic energy per planet.
or make them ... half of 1.3x10^11 and the other just 6.99x10^7 ?
Why would you do that?
well...I i know i get the right answer from using 1.3x10^11 and (2*6.99x^10^7) but what is the reason to use the diameter instead of cutting them in half ...?
What's important is the distance between their centers. When they collide, that distance equals twice the radius. Right?

also, is that the right way to set it up because I get the right answer from my 42,556 / sqrt (2)...or is that the same thing by making 2K
Work it out and see! :smile:
 
  • #10
Thanks for all your help. I have a really good grasp on it now.
 

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